1 Description #
source: https://leetcode.com/problems/merge-intervals/
Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:
Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
2 Solution #
from typing import List
# time complexity: O(nlogn), time to sort intervals.
# space complexity: O(1)
class Solution:
def merge(self, intervals: List[List[int]]) -> List[List[int]]:
size = len(intervals)
if size <=1:
return intervals
# sorted
merged = []
intervals.sort(key = lambda x: x[0])
for interval in intervals:
if not merged or merged[-1][1]<interval[0]:
merged.append(interval)
else:
merged[-1][1] = max(merged[-1][1], interval[1])
return merged
#include <algorithm>
class Solution {
public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
std::sort(intervals.begin(), intervals.end());
std::vector<std::vector<int>> merged;
for(const auto& interval: intervals){
if(merged.empty() || merged.back()[1] < interval[0]){
merged.push_back(interval);
}else{
// there is an overlap
merged.back()[1] = std::max(merged.back()[1], interval[1]);
}
}
return merged;
}
};